Problem: Simplify and expand the following expression: $ \dfrac{q}{q - 4}-\dfrac{4q + 10}{q - 8} $
Solution: In order to subtract expressions, they must have a common denominator. Get both fractions over a common denominator of $(q - 4)(q - 8)$ Multiply the first term by $\dfrac{q - 8}{q - 8}$ $ \begin{align*} \dfrac{q}{q - 4} \times \dfrac{q - 8}{q - 8} & = \dfrac{(q)(q - 8)}{(q - 4)(q - 8)} \\ & = \dfrac{q^2 - 8q}{(q - 4)(q - 8)}\end{align*} $ Multiply the second term by $\dfrac{q - 4}{q - 4}$ $ \begin{align*} \dfrac{4q + 10}{q - 8} \times \dfrac{q - 4}{q - 4} & = \dfrac{(4q + 10)(q - 4)}{(q - 8)(q - 4)} \\ & = \dfrac{4q^2 - 6q - 40}{(q - 8)(q - 4)}\end{align*} $ Now we have: $ = \dfrac{q^2 - 8q}{(q - 4)(q - 8)} - \dfrac{4q^2 - 6q - 40}{(q - 8)(q - 4)} $ Now both terms have a common denominator we can subtract the numerators: $ = \dfrac{q^2 - 8q - (4q^2 - 6q - 40)}{(q - 4)(q - 8)} $ $ = \dfrac{q^2 - 8q - 4q^2 + 6q + 40}{(q - 4)(q - 8)} $ $ = \dfrac{-3q^2 - 2q + 40}{(q - 4)(q - 8)}$ Expand the denominator: $ = \dfrac{-3q^2 - 2q + 40}{q^2 - 12q + 32}$